Question

A box

of mass 500 g is moving with a speed of

2.5 ms-1on a table. It comes to a stop

after travelling a distance of

2.6 m. Calculate the coefficient of

kinetic friction between the box and the

table. Take g = 10 ms 2.

Answer 1

mass=0.5kg
speed=2m/s
distance=2.6m

use eq.
${v}^{2} - {u}^{2} = 2as$
v=0

-(2.5)^2=2(-a)2.6

a=1.20m/s^2


uN=ma
u×(mg)=ma
ug=a
u=1.20/10
u=0.12

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